The Distance Formula

Plot $(-3, 2)$ and $(5, 8)$ and join them. Now add a horizontal line and a vertical line to make a right triangle with that interval as the hypotenuse. How long is each of the two legs? What theorem finishes the job?

Complete the right triangle under the interval. Its legs run parallel to the axes, so their lengths are the coordinate differences. Then Pythagoras theorem turns those two numbers into the length of the interval itself.

Two points on the plane are joined by an interval that is, in general, slanted. Slanted lengths are hard to measure directly, but horizontal and vertical ones are easy, because you can read them off the coordinates.

So build a triangle that has one of each. From the left-hand point draw a horizontal line, from the right-hand point draw a vertical one, and let them meet. The interval becomes the hypotenuse of a right triangle whose legs are parallel to the axes.

You have converted an unmeasurable length into two measurable ones plus a theorem. That conversion is the entire idea, and the algebra that follows is bookkeeping.

The horizontal leg runs from $x_1$ to $x_2$, so its length is $x_2 - x_1$. The vertical leg runs from $y_1$ to $y_2$, so its length is $y_2 - y_1$.

Strictly, a length cannot be negative, so those should be written with absolute value signs as $|x_2 - x_1|$ and $|y_2 - y_1|$. Pythagoras theorem then gives

$d^2 = |x_2 - x_1|^2 + |y_2 - y_1|^2$

But squaring a number and squaring its negative give the same result, so the absolute value signs do nothing and can be dropped:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

That is why the formula looks tidier than the geometry it came from. The squaring absorbs the sign problem, which is a small piece of luck worth noticing rather than passing over.

In the gradient formula the order of the points was safe only if you were consistent, because reversing just one subtraction flipped the sign. Distance has no such trap.

Reverse the points and each difference becomes its own negative, but each is then squared, and squaring destroys the sign entirely. $(5 - (-3))^2 = 8^2 = 64$ and $(-3 - 5)^2 = (-8)^2 = 64$.

So you may subtract in whichever order you find easier, even inconsistently between the two brackets, and the answer is unaffected.

What still matters is where the square root sits. It covers the whole sum. Writing $\sqrt{(x_2 - x_1)^2} + \sqrt{(y_2 - y_1)^2}$ computes the two legs and adds them, which is the distance a taxi would drive, not the distance a drone would fly.

Most distances are irrational, so the honest answer is a surd.

Simplify it where a square factor allows. $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$, and $\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$. Where no square factor exists, as in $\sqrt{74}$, the surd is already in simplest form and is a complete answer.

Round only when the question asks, or when the context demands it. A distance on a map is naturally a decimal; a distance inside a proof should stay exact, because rounding early makes later comparisons unreliable.

That last point matters for the next card. Deciding whether two sides are equal is trivial with exact values and genuinely awkward with rounded ones, since $\sqrt{50}$ and $\sqrt{50.01}$ both round to $7.07$.

Given three vertices, three applications of the formula give the three side lengths, and the lengths classify the triangle.

All three different means scalene. Exactly two equal means isosceles. All three equal means equilateral.

To test for a right angle, use the converse of Pythagoras theorem. Identify the longest side, square it, and compare with the sum of the squares of the other two. If they are equal the triangle is right-angled, with the right angle opposite the longest side. If they are not equal, it is not right-angled.

Working with squared lengths makes this painless, since the squares are what the formula produces before the final root. A triangle can be both isosceles and right-angled, and that combination shows up often enough to be worth expecting rather than doubting.

Five distance items. Leave every answer exact, simplifying the surd where possible.

1. 1 Find the distance between $(0, 0)$ and $(6, 8)$.

   Show answer
   $\sqrt{36 + 64}$$10$

2. 2 Find the distance between $(-1, -1)$ and $(2, 3)$.

   Show answer
   Differences of $3$ and $4$.$5$

3. 3 Find the distance between $(2, 5)$ and $(2, -3)$.

   Show answer
   A vertical interval, so the horizontal difference is zero.$8$

4. 4 Find the distance between $(-3, 4)$ and $(3, -4)$.

   Show answer
   $\sqrt{6^2 + 8^2}$$10$

5. 5 Find the exact distance between $(1, 2)$ and $(5, 9)$.

   Show answer
   $\sqrt{16 + 49}$, and $65 = 5 \times 13$ has no square factor.$\sqrt{65}$